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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Given a function <span class="process-math">\(f,\)</span> of <span class="process-math">\(t,\)</span> we denote its Laplace Transform by <span class="process-math">\({\mathcal L}[f(t)]=F(s);\)</span> the inverse process is written:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}^{-1}[F(s)]=f(t).
\end{equation*}
</div>
<p class="continuation">Frequently, a Laplace transform <span class="process-math">\(F(s)\)</span> is expressible as a sum of several terms,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
F(s)=F_1(s)+F_2(s)+\cdots+F_n(s).
\end{equation*}
</div>
<p class="continuation">Suppose that <span class="process-math">\(f_1(t)={\mathcal L}^{-1}\{F_1(s)\}, \cdots, f_n(t)={\mathcal L}^{-1}\{F_n(s)\}\text{.}\)</span> Then the function</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f(t)=f_1(t)+f_2(t)+\cdots f_n(t)
\end{equation*}
</div>
<p class="continuation">has the Laplace transform <span class="process-math">\(F(s)\text{.}\)</span> By the uniqueness property, no other continuous function <span class="process-math">\(f\)</span> having the same transform. Thus</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}^{-1}[F(s)]={\mathcal L}^{-1}[F_1(s)]+{\mathcal L}^{-1}[F_2(s)]+\cdots+{\mathcal L}^{-1}[F_n(s)],
\end{equation*}
</div>
<p class="continuation">that is, the inverse Laplace transform is also a linear operator.</p>
<span class="incontext"><a href="sec8_3.html#p-449" class="internal">in-context</a></span>
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